Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a way to define a differentiable function in a non-pointwise manner? That is without defining function differentiable at a point first.

Just like we define a continuous function through open subsets.

As far as I recall there was something involving the chain rule and probably Leibniz rule, but I don't remember by now.

share|cite|improve this question

I don't know if this is what you are looking for, but here are some thoughts.

Differentiability is defined at a point, however, there are some concepts which relate differentiability to more global properties. The first that come to my mind are the Fourier Transform and Lipschitz Continuity.

The Fourier Transform of the derivative of a function at $\xi$ is the Fourier Transform of that function times $\xi$; that is, $\widehat{f'}(\xi)=2\pi i\xi\widehat{f}(\xi)$. In this way, the differentiability of a function is related to decay of its Fourier Transform. The smoother a function is, the faster its Fourier Transform decays near $\infty$.

The idea of Lipschitz Continuity relates a kind of uniform smoothness to the $L^\infty$ norm of the difference of a function and nearby translates. That is, $\|f-T_hf\|_{L^\infty}\le C|h|$ where $T_hf(x)=f(x+h)$ is a translation operator. In this case, $f$ is differentiable almost everywhere and $\|f'\|_{L^\infty}\le C$.

share|cite|improve this answer

Here is another one:

A function $f:\ I\to{\mathbb R}$ is continuously differentiable on the open interval $I$ iff there is a continuous function $L:\ I\times I\to{\mathbb R}$ such that $$f(y)-f(x)=L(x,y)\,(y-x)\qquad (x, y\in I)\ .$$ Proof. If $f\in C^1(I)$ then $$f(y)-f(x)=\int_x^y f'(t)\ dt=(y-x)\int_0^1 f'\bigl((1-\tau)x+\tau y\bigr)\ d\tau\ ;$$ therefore the function $$L(x,y)\ :=\ \int_0^1 f'\bigl((1-\tau)x+\tau y\bigr)\ d\tau$$ has the required properties. Conversely, if there is such an $L$ then $$f'(x)=\lim_{y\to x}{f(y)-f(x)\over y-x}=\lim_{y\to x}L(x,y)=L(x,x)\ ,$$ and the right side is continuous as a function of $x$.

This principle can be extended to the multivariable case.

share|cite|improve this answer
This is certainly correct and if you use the "global" definition of continuity mentioned by the OP you get what is in some sense a "global" definition of (continuous) differentiability. But doesn't this feel somewhat "cheap"? It seems you have placed very thin wrapping paper around the usual definition of differentiability... – Pete L. Clark Mar 18 '12 at 18:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.